Let ${x_k}={2^{(-1)}}^{k}(1+1/k)$
Find $\liminf x_k$ and $\limsup x_k$.
I tried in this way.
First, I split {$x_k$} into two subsequences; $x_{2j}=2+\frac{1}{j}$ and $x_{2j-1}=\frac{1}{2}+\frac{1}{4j-2}$
Then {$x_{2j}$} is monotonically decreasing from max 3 to a limit 2,
and {$x_{2j-1}$} is monotonically decreasing from max 1 to a limit 1/2.
It follows that $\frac{1}{2}\le x_k\le 3$, thus $\frac{1}{2}\le \liminf x_k \le \limsup x_k \le 3$
Then by picking {$x_{2j}$}, we can verify that $\lim_{j\to \infty}x_{2j}=2$
and by picking {$x_{2j-1}$} we can verify that $\lim_{j\to \infty}x_{2j-1}=\frac{1}{2}$
My TA gave me one point out of five to this answer.
I have tried to figure out why and my guess is
maybe because {$x_{2j}$} and {$x_{2j-1}$} are not the only subsequences that I can choose from {$x_{k}$}.
Or because I did not use the definition of cluster point.
Please give me some help!
(1) $x_{2j}=2+\frac{1}{4j}$. That does not affect the answer much, but it is still an error.
(2) $x_{2j-1}=-(\frac{1}{2}+\frac{1}{4j-2})$. You forgot the minus sign. That does make a difference. You end up with $\frac{1}{2}$ instead of $-\frac{1}{2}$. Also the subsequence is increasing not decreasing.
(3) you do not explicitly deal with the fact that you are asked for lim sup and lim inf. So you need to explain why you have got the largest and smallest limit points.